Question

On the interval $[0,1]$ the function $f\left(x\right)=x^{1005}(1-x{)}^{1002}$ assumes maximum value equal to.

• A. $\frac{\left(1005{)}^{1002}\right(1002{)}^{1005}}{(2007{)}^{2007}}$
• B. $\frac{(2007{)}^{2007}}{\left(1005{)}^{1005}\right(1002{)}^{1002}}$
• C. $\frac{(2007{)}^{2007}}{\left(1005{)}^{1002}\right(1002{)}^{1005}}$
• D. $\frac{\left(1005{)}^{1005}\right(1002{)}^{1002}}{(2007{)}^{2007}}$

Answer 1

The correct option is D $\frac{\left(1005{)}^{1005}\right(1002{)}^{1002}}{(2007{)}^{2007}}$

$f\left(x\right)=x^{1005}(1-x{)}^{1002}$

We know that for a function $f\left(x\right)$ at the point where $f^{\prime }\left(x\right)=0f\left(x\right)$ assumes its maximum and minimum values.

$f^{\prime }\left(x\right)=x^{1005}\left(1002\right)(1-x{)}^{1001}(-1)+(1-x{)}^{1002}1005x^{1004}=x^{1004}(1-x{)}^{1001}[1005-1005x-1002x]$

$f^{\prime }\left(x\right)=x^{1004}(1-x{)}^{1001}[1005-2007x]$

$f^{\prime }\left(x\right)=0$

$x=0,x=1,x=\frac{1005}{2007}$

$f\left(x\right)$ assumes its maximum value at $x=\frac{1005}{2007}$

$f(x{)}_{x=\frac{1005}{2007}}={\left(\frac{1005}{2007}\right)}^{1005}{(1-\frac{1005}{2007})}^{1002}=\frac{{\left(1005\right)}^{1005}{\left(1002\right)}^{1002}}{{\left(2007\right)}^{2007}}$